[seqfan] Re: formula for Pi

[Robert Israel]

This series has no simple relation to the digits of Pi (whether base 
10 or any other base).  Moreover, it converges quite slowly, since the 
terms are of order 1/n^2.  By the way, it follows quite easily from the 
identity

sum_{n=0}^infty (1/(n+b) - 1/(n+a)) = Psi(a) - Psi(b)

together with
Psi(5) = 25/12-gamma,
Psi(22/3) = 699171/138320-gamma-1/6*Pi*3^(1/2)-3/2*ln(3),
Psi(7/3) = 15/4-gamma-1/6*Pi*3^(1/2)-3/2*ln(3)

noting that the summand is 22/(n+7/3) - 7/(n+5) - 15/(n+22/3).
A nicer version, in my opinion, is

sum_{n=0}^infty 1/((3*n+1)*(3*n+2)) = Pi*sqrt(3)/9

Cheers,
Robert Israel

On Sun, 7 Dec 2008, Hector Zenil wrote:

> Is it possible to get subsequences of the digits of Pi from this
> formula? Of course without doing N[formula,digits] or any other
> similar trick...
>
>
> On Sun, Dec 7, 2008 at 3:28 PM, Vladimir Bondarenko <vb at cybertester.com> wrote:
>> Hello,
>>
>> This is an exact formula for Pi.
>>
>> FullSimplify[6/7*(1/3*Sum[(843*n + 4607)/((n + 5)*(3*n + 7)*(3*n + 22)),
>> {n, 0, Infinity}] - 655999/248976 - 7/2*Log[3])*Sqrt[3]]
>>
>> Pi
>>
>> Cheers,
>>
>> Vladimir Bondarenko
>>
>> Quoting Alexander Povolotsky <apovolot at gmail.com>:
>>
>>> I've got this very ugly formula by playing Maple syntax via old and
>>> new inverse symbolic calculators :
>>>
>>> Pi = 6/7*(1/3*sum((843*n + 4607)/((n+5)*(3*n+7)*(3*n+22)),n=0...infinity)
>>>  - 655999/248976  - 7/2*ln(3))*sqrt(3)
>>>
>>> What I've got for Pi above - is it just a good approximation or exact ?
>>> (My old PC with PARI/GP can not get over the summing )
>>>
>>> It looks that this 655999/248976 fraction quickly becomes periodic around
>>>  920249341301972880
>>>
>>> gp > \p 1000
>>> realprecision = 1001 significant digits (1000 digits displayed)
>>> gp > 1.0*655999/248976
>>> %4 = 2.6347880
>>> 920249341301972880
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>>> 92024934130197......
>>> 288092....
>>>
>>> Cheers Alexander R. Povolotsky
>>>
>>>
>>> _______________________________________________
>>>
>>> Seqfan Mailing list - http://list.seqfan.eu/
>>>
>>
>>
>>
>>
>>
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>>
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>
>
>
> -- 
> Hector Zenil				http://www.mathrix.org
>
>
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